# Evaluation of an Infinite Sum?

• Nov 28th 2006, 10:38 AM
TMFKAN64
Evaluation of an Infinite Sum?
I'm looking to evaluate the infinite sum of 1/(n-f)^2 where n is the index of summation from 1 to infinity and f is a constant between zero and one.

I really have no clue where to start on this. I've seen the summation evaluated to pi^2/6 if f=0 by using a Fourier transform of x^2, but I can't see how to adapt this for an arbitrary constant.

Any ideas or hints would be appreciated.
• Nov 28th 2006, 10:53 AM
ThePerfectHacker
Quote:

Originally Posted by TMFKAN64
I'm looking to evaluate the infinite sum of 1/(n-f)^2 where n is the index of summation from 1 to infinity and f is a constant between zero and one.

I really have no clue where to start on this. I've seen the summation evaluated to pi^2/6 if f=0 by using a Fourier transform of x^2, but I can't see how to adapt this for an arbitrary constant.

Any ideas or hints would be appreciated.

Maybe, ... it would seem reasonable to find the Fourier series for the function $\displaystyle f(x)=(x-f)^2$ defined on $\displaystyle [-\pi,\pi]$ and with periodic property $\displaystyle f(x+2\pi)=f(x)$.
• Nov 28th 2006, 02:01 PM
TMFKAN64
That was my initial thought, but it doesn't really help. In the Fourier transform method when f=0, we end up finding that x^2 = pi^2/3 + 4 sum (-1)^n/n^2 cos nx. This becomes our sum when we let x=pi. If I try (x-f)^2 instead, I end up with some sin terms and a different constant offset, but the denominator is still n^2. (I get (x-f)^2 = pi^2/3 + f + 4 sum [(-1)^n/n^2 cos nx + f pi/n sin nx as the fourier series.)
• Nov 28th 2006, 05:12 PM
ThePerfectHacker
Quote:

Originally Posted by TMFKAN64
That was my initial thought, but it doesn't really help. In the Fourier transform method when f=0, we end up finding that x^2 = pi^2/3 + 4 sum (-1)^n/n^2 cos nx. This becomes our sum when we let x=pi. If I try (x-f)^2 instead, I end up with some sin terms and a different constant offset, but the denominator is still n^2. (I get (x-f)^2 = pi^2/3 + f + 4 sum [(-1)^n/n^2 cos nx + f pi/n sin nx as the fourier series.)

First it is more correct to say,
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(n-f)^2}$
Is undefined if $\displaystyle f$ is an integer.
So we should use a plus sign and state that $\displaystyle f$ is a positive integers or non-integral real number.
Note that if $\displaystyle f$ is an positive integer then you get the standard "Basel problem" only shifted over.